The other line extends slightly down and to the right to point E
For this problem, you would have to use a system of equations. Hope this helps!
You want the (positive) distance when the height is 0, 0= -d^2 + 10d +5.
You would factor that out(You'd get radicals), and the final answer would be
<span>5±<span>30<span>−−</span>√</span></span><span>
You'd take the only positive one of the two, 5 + srt(30)
I hope I did that right :/</span>
Circle: x^2+y^2=121=11^2 => circle with radius 11 and centred on origin.
g(x)=-2x+12 (from given table, find slope and y-intercept)
We can see from the graphics that g(x) will be almost tangent to the circle at (0,11), and that both intersection points will be at x>=11.
To show that this is the case,
substitute g(x) into the circle
x^2+(-2x+12)^2=121
x^2+4x^2-2*2*12x+144-121=0
5x^2-48x+23=0
Solve using the quadratic formula,
x=(48 ± √ (48^2-4*5*23) )/10
=0.5058 or 9.0942
So both solutions are real and both have positive x-values.
